spfa版费用流：
1 //最小费用最大流spfa版，求最大费用只需要取相反数，结果取相反数即可。 2 //点的编号0～n-1 3 const int MAXN = 10000; 4 const int MAXM = 100000; 5 const int INF = 0x3f3f3f3f; 6 struct Edge 7 { 8 int to, next, cap, flow, cost; 9 }edge[MAXM]; 10 int head[MAXN], tol; 11 int pre[MAXN], dis[MAXN]; 12 bool vis[MAXN]; 13 int N; 14 void init(int n) 15 { 16 N = n; 17 tol = 0; 18 memset(head, -1, sizeof(head)); 19 } 20 void addedge(int u, int v, int cap, int cost) 21 { 22 edge[tol].to = v; edge[tol].cap = cap; 23 edge[tol].cost = cost; edge[tol].flow = 0; 24 edge[tol].next = head[u]; head[u] = tol++; 25 edge[tol].to = u; edge[tol].cap = 0; 26 edge[tol].cost = -cost; edge[tol].flow = 0; 27 edge[tol].next = head[v]; head[v] = tol++; 28 } 29 bool spfa(int s, int t) 30 { 31 queue<int> q; 32 for(int i = 0; i < N; i++) 33 { 34 dis[i] = INF; 35 vis[i] = false; 36 pre[i] = -1; 37 } 38 dis[s] = 0; vis[s] = true; 39 q.push(s); 40 while(!q.empty()) 41 { 42 int u = q.front(); 43 q.pop(); 44 vis[u] = false; 45 for(int i = head[u]; i != -1; i = edge[i].next) 46 { 47 int v = edge[i].to; 48 if(edge[i].cap > edge[i].flow && dis[v] > dis[u] + edge[i].cost) 49 { 50 dis[v] = dis[u] + edge[i].cost; 51 pre[v] = i; 52 if(!vis[v]) 53 { 54 vis[v] = true; 55 q.push(v); 56 } 57 } 58 } 59 } 60 if(pre[t] == -1) return false; 61 else return true; 62 } 63 //返回的是最大流，cost存的是最小费用 64 int minCostMaxflow(int s, int t, int &cost) 65 { 66 int flow = 0; 67 cost = 0; 68 while(spfa(s, t)) 69 { 70 int Min = INF; 71 for(int i = pre[t]; i != -1; i = pre[edge[i^1].to]) 72 { 73 if(Min > edge[i].cap - edge[i].flow) 74 Min = edge[i].cap - edge[i].flow; 75 } 76 for(int i = pre[t]; i != -1; i = pre[edge[i^1].to]) 77 { 78 edge[i].flow += Min; 79 edge[i^1].flow -= Min; 80 cost += edge[i].cost * Min; 81 } 82 flow += Min; 83 } 84 return flow; 85 }